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2 years ago

Ed is 7 years older than Ted. Ed’s age is also 3/4 times Ted’s age. How old are Ed and Ted?

1 answer:

7 0

Let Ted be x.

Ed is 7 years older = x + 7

Ed = (3/4)Ted

(x + 7) = (3/4)x

x + 7 = 3x/4

x - 3x/4 = -7

x/4 = -7

x = -28, Ted = -28 years.

(x + 7) = -28 + 7 = -21, Ed = -21 years

Goodness. We had negative numbers for the ages, well does that make sense? No it doesn't.

Our answer is correct. But the sense in the question is lacking. The question has been wrongly set.

We might assume negative ages to mean before they came into the world, before birth!

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h----> the height of the triangle

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2 years ago

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Answer: 1/10

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1 year ago

If two lines are perpendicular describe the relationship between their slopes

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Perpendicular lines have slopes that are negative reciprocals. example: line a has a slope of 2/3, line b has a slope if -3/2 if they are perpendicular.

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2 years ago

A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a st

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Answer:

95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

Step-by-step explanation:

We are given that a random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample credit hours per quarter = 15.2 credit hours

s = sample standard deviation = 2.3 credit hours

n = sample of students = 250

$\mu$ = population mean credit hours per quarter

Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < $t_2_4_9$ < 1.96) = 0.95 {As the critical value of t at 249 degree of

freedom are -1.96 & 1.96 with P = 2.5%}

P(-1.96 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $15.2-1.96 \times {\frac{2.3}{\sqrt{250} } }$ , $15.2+1.96 \times {\frac{2.3}{\sqrt{250} } }$ ]

= [14.915 hours , 15.485 hours]

Therefore, 95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

The interpretation of the above confidence interval is that we are 95% confident that the true mean credit hours taken by a student each quarter will be between 14.915 credit hours and 15.485 credit hours.

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2 years ago

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Nostrana [21]

If the triangles are congruent, the angle opposite the 32 cm side will have the same measure as angle E, 40°. An appropriate choice for this question is ...
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Even if you overlook the units problem, the sum of squares of the sides of the triangle is 32² +38² = 2468, somewhat short of 50² = 2500. That is, the largest angle is slightly more than 90°. It is closer to 90.754° ≈ 91°.

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